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Chapter 3: Problem 73

Each copper(II) sulfate unit is associated with five water molecules in crystalline copper(II) sulfate pentahydrate \(\left(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\right)\). When this compound is heated in air above \(100^{\circ} \mathrm{C},\) it loses the water molecules and also its blue color: $$ \mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{CuSO}_{4}+5 \mathrm{H}_{2} \mathrm{O} $$ If \(9.60 \mathrm{~g}\) of \(\mathrm{CuSO}_{4}\) are left after heating \(15.01 \mathrm{~g}\) of the blue compound, calculate the number of moles of \(\mathrm{H}_{2} \mathrm{O}\) originally present in the compound.

Short Answer

Expert verified
Thus, the number of moles of \(H_{2}O\) originally present in the compound was 0.3 moles.

Step by step solution

01

Determine the mass of water

First, one can find the mass of the water that was in the original hydrate. This can be done by subtracting the mass of the copper(II) sulfate that remains after heating from the mass of the original compound. In mathematical terms: mass of water = mass of original compound - mass of copper(II) sulfate. \[ mass_{H2O} = 15.01g - 9.60g = 5.41g \]
02

Calculate the Number of Moles of Water

With the mass of water obtained, calculate the number of moles. Moles can be calculated by dividing the mass of the compound by its molar mass. For water, the molar mass is 18.016 g/mol. Hence, \[ moles_{H2O} = \frac{mass_{H2O}}{18.016 g/mol} \]
03

Perform the Calculation

Applying the numbers from above: \[ moles_{H2O} = \frac{5.41g}{18.016 g/mol} = 0.3 mol \] The result shows the number of moles of water originally present in the compound.

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Most popular questions from this chapter

A certain metal M forms a bromide containing 53.79 percent Br by mass. What is the chemical formula of the compound?

What are the empirical formulas of the compounds with the following compositions? (a) 40.1 percent \(\mathrm{C}\) 6.6 percent \(\mathrm{H}, 53.3\) percent \(\mathrm{O}\) (b) 18.4 percent \(\mathrm{C}\) 21.5 percent \(\mathrm{N}, 60.1\) percent \(\mathrm{K}\)

The aluminum sulfate hydrate \(\left[\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} \cdot x \mathrm{H}_{2} \mathrm{O}\right]\) contains 8.20 percent Al by mass. Calculate \(x\), that is, the number of water molecules associated with each \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) unit.

Give an everyday example that illustrates the limiting reagent concept.

A sample of iron weighing \(15.0 \mathrm{~g}\) was heated with potassium chlorate \(\left(\mathrm{K} \mathrm{ClO}_{3}\right)\) in an evacuated container. The oxygen generated from the decomposition of \(\mathrm{KClO}_{3}\) converted some of the Fe to \(\mathrm{Fe}_{2} \mathrm{O}_{3}\). If the combined mass of Fe and \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) was \(17.9 \mathrm{~g},\) calculate the mass of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) formed and the mass of \(\mathrm{KClO}_{3}\) decomposed.
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